Modeling Disease Spread: Differential Equation for Infected Individuals

[Natasha1]

The spread of a disease in a community is modeled by the following differential equation:

dy/dx = 0.2y - 0.02x where y is the number of infected individuals in thousands, and x the time in days.

1) Show the equation that for the family of 'curves' in the x-y plane for which dy/dx is a constant, is of the form y=mx+c.

How can I do this please?

 

[AKG]

You have the differential equation:

y' - 0.2y = p(x)

where p(x) is some polynomial in x, and y' denotes dy/dx. Find the family of functions that solves this equation, and determine the one that has y' constant.

 

[Natasha1]

You have the differential equation:

y' - 0.2y = p(x)

where p(x) is some polynomial in x, and y' denotes dy/dx. Find the family of functions that solves this equation, and determine the one that has y' constant.

p(x) = -0.02x no?

 

[Natasha1]

How is this done :-(

 

[AKG]

Yes, that's correct for p(x). To find the general solution y, a function of x, to this problem, you find

i) the general solution to the related homogeneous equation y' - 0.2y = 0
ii) find anyone particular solution to the original homogenous equation
iii) add them

So finding the general solution to y' - 0.2y = 0 means finding a family of solutions to this equation. We can do it:

y' - 0.2y = 0
y' = 0.2y
dy/dx = 0.2y
dy/y = 0.2dx
lny = 0.2x + C
y = exp(0.2x + C)
y = a*exp(0.2x) [where a = e^C]

So this gives a family of solutions, paramterized by a. In other words, for EACH real number a, y(x) = ae^0.2x is a solution to y' - 0.2y = 0. We write y_h(x) = ae^0.2x, where the subscript "h" stands for "homogenous"

We now look for y_p, a particular solution to the original equation:

y' - 0.2y = p(x)

Since p(x) is a degree 1 polynomial, it's a rule of thumb that we ought to guess a degree 1 polynomial for our particular solution. Well a degree 1 polynomial is just one of the form mx+b, so we let this be our guess of y_p, and we sub it into see if it works, and what values of m, b we need:

If y_p = mx+b, then y_p' = m, so:

y_p' - 0.2y_p = -0.02x
m - 0.2(mx + b) = -0.02x
-0.2mx + (m - 0.2b) = -0.02x + 0

So -0.2m = -0.02 [equating coefficients of x] and m-0.2b = 0 [equating coefficients of 1]

We get m = 1/10, b = 1/2, giving y_p = 0.1x + 0.5

So the general solution is y = y_h + y_p = ae^0.2x + 0.1x + 0.5. What this means is that a function y of x is a solution to dy/dx = 0.2y - 0.02x if and only if there is some real number a such that y(x) = ae^0.2x + 0.1x + 0.5. So if every solution comes in the form ae^0.2x + 0.1x + 0.5, which of these have derivative constant?

y' = 0.2ae^0.2x + 0.1

Well this is constant iff a = 0, correct? And a = 0 iff y = 0.1x + 0.5. So y' is constant iff y = 0.1x + 0.5 which implies that y is in the form y = mx+b.

 

[nrqed]

The spread of a disease in a community is modeled by the following differential equation:

dy/dx = 0.2y - 0.02x where y is the number of infected individuals in thousands, and x the time in days.

1) Show the equation that for the family of 'curves' in the x-y plane for which dy/dx is a constant, is of the form y=mx+c.

How can I do this please?

dy/dx is a constant, call it ''k'' let's say. Then replace dy/dx in your initial equation by k and isolate y.

 

[AKG]

Yes, do what nrqed said and totally ignore what I said. It's way more work than necessary.

 

[Natasha1]

Yes, do what nrqed said and totally ignore what I said. It's way more work than necessary.

so

0.2y = k + 0.02x

Hence

y = 0.1x + 5k is this correct?

 

[AKG]

Yes...